From dcaea06cc0e3617dbaa39c86e89cebb129fedc01 Mon Sep 17 00:00:00 2001 From: Barrett Ruth Date: Thu, 10 Apr 2025 12:43:58 -0400 Subject: [PATCH] feat(codeforces): 779 --- .../competitive-programming-log.html | 101 +++++++++++++++++- 1 file changed, 100 insertions(+), 1 deletion(-) diff --git a/posts/algorithms/competitive-programming-log.html b/posts/algorithms/competitive-programming-log.html index 74eadc1..319d5f7 100644 --- a/posts/algorithms/competitive-programming-log.html +++ b/posts/algorithms/competitive-programming-log.html @@ -35,6 +35,104 @@

Competitive Programming Log

+

+ 799 (div. 4)—10/4/2025 +

+
+ Improvement is marginal. My desire for rating is unquenchable + (joke). Really, though, I'm improving slower than I like. 1400 + performance, my best yet—I'll do one more Div. 4 then I need + to upgrade to Div. {2,3}. I think the most core realization I made + after this contest was: +
+ Separate what your code does from what you think it should do. + Conceptualize an approach, then put it aside. Implement exactly + what the approach says to do (not what you think it should say). + If the approach left something out, + step away from implementing, reconsider the approach, then + resume implementing. Never reconsider and alter your strategy while coding. +
+
+
    +
  1. + B: realized parity-based removal immediately. Still, I once again + didn't initially do what the problem asked (i.e. I returned the + number of operations, not array length). +
    2. +
    + Never run code until you're confident it does what you want. +
    +
  2. + C: Blindly plugged in incorrect deltas to detect the "X" pattern + of "#" characters before getting the right answer. +
    + I consistently tune out when I'm bored/think I've discovered the + insight/done the problem before. This results in me solving a + different problem, expressing right ideas incorrectly, or + expressing wrong ideas correctly—for example, thinking I + find a core insight, believing it sounds right, and just + not checking its validity. I don't know what to say + besides progressively overload your discipline and focus. +
    +
    3. +
  3. + D: my proof that the time must repeat was complete garbage and + complex. I've been reading AOPS and wanted to go with a math-y + approach. However, consider the following simple proof: +
    + The current time is \(s\) and you look at the clock every \(x\) + minutes. \(1440 \cdot x\equiv 0\pmod{1400}\), so the time must + repeat after at most 1440 looks. +
    + This is enough to brute force. There are alternative easier + approaches to brute-force, but for an easy problem like this speed + comes above all else (special case). +
    4. +
  4. + E: Submitted and hoped for correctness multiple times. I + was lazy and didn't want to implement the binary search approach. + Ironically, after turning my brain on, I quickly found a + \(\Theta(n)\) solution. However, I rushed to code it up thinking + about the time I had wasted and frustration from WA (more from + myself than from WA). This caused me to forget invariants and + implement what I thought was right, not the algorithm I designed + (forgetting invariants). Walking through an example (as I advised + myself yesterday) reminded me to iterate the right pointer past + any zeroes. Good, specific improvement from yesterday, but I still + should've slowed down and asked "where should the right pointer + move? what are the invariants? why? what about if if goes off the + end of the array?" Only look forward. +
    5. +
  5. + F: Came up with fancy equations because I'm taking the wrong core + lessons from AOPS. The example problems there are complex, while + these are simple—therefore, I should prioritize simplicity. + Concretely, the following phrasing massively simplifies this + problem: +
    + I'm looking for 3 numbers \(a[i],a[j],a[k], i\neq j\neq k\) such + that \(a[i]+a[j]+a[k]\equiv 3\pmod{10}\). WLOG, assume \(i\lt + j,k\). We only want a pair of remainders that satisfy the above + inequality. Since there are only 10 unique remainders, brute + force all 100 remaining modulo pairs. If such a pair exists in + \(a[i+1:]\forall\) \(i\in\{1,...n-2\}\), the answer is yes. This + can be calculated with a postfix frequency map of remainders, + with special care taken for when the remainders are equal. +
    +
    6. +
  6. + G: good deduction and simplicity but I once again read the + instructions too fast and skipped that the array is size \(k+1\). + Also, + my sliding window implementation ability is horrendous. I + spent nearly 10 minutes just debugging a "right" idea because I + couldn't code up a basic thing. +
    7. +

849 (div. 4)Did not read the problem statement and answered a similar (but very different) problem I'd done in the past. By the time I saw this after impatiently submitting ~5 WA, I - had lost mental focus. Never submit and hope for a correct answer—know it. + had lost mental focus. Never submit and hope for a correct + answer—know it.
  • F: Then, I let my previous failure carry through to the next